Part IV.
i) a) Moles of CH3COONa = 4.23 g/(82 g/mol) = 0.0515854 mol
Moles of CH3COOH = (8.5 mL * 1.05 g/mL)/(60 g/mol) = 0.14875 mol
According to the Henderson-Hasselbulch equation:
pH = pKa + Log([CH3COONa]/[CH3COOH])
i.e. Calculate pH for the buffer = 4.74 + Log(0.0515854/0.14875) = 4.28
b) calculated pH after adding HCl to buffer = 4.74 + Log{(0.0515854-0.006)/(0.14875+0.006)} = 4.21
c) calculated pH after adding NaOH to buffer = 4.74 + Log{(0.0515854+0.006)/(0.14875-0.006)} = 4.35
ii) a) Calculate pH for the DI water = -Log[H+] = -Log(10-7) = 7
b) calculated pH after adding HCl to DI water = -Log(1*6/41) = 0.835
c) calculated pH after adding NaOH to DI water = 14-{-Log(1*6/41)} = 13.165
The notable thing is that the change in pH after the addition of HCl solution and NaOH solution for the buffer can be neglected.
Explanation: A buffer solution resists the change in pH by the addition of a small amount of strong acid or strong base.
Whereas the change in pH after the addition of the HCl solution and NaOH solution for DI water is much more significant.
Explanation: DI water as such is neutral (pH = 7); when you add strong acid to it, it will be acidic (pH < 7), when you add a strong base to it, it will be basic (pH > 7)
thats all i'm given would it be the Ka1 which js 1.8x10^-5 pH and Buffers IV....
pH and Buffers IV. pH of a Buffer Solution Mass of sodium acetate, CH,COONa, g 4.2009 COOM Calculated pH Measured pH buffer solution prepared with dissolved CHCOONa+ 8.5 mL CH, COOH(aq) 5.54 40 mL buffer + 1.0 mL of 6.0 M HCl(aq) 14.05 40 mL buffer + 1.0 mL of 6.0 M NaOH(aq) 5.02 Calculated pH Measured pH 5.54 1063 deionized water 40 mL DI water + 1.0 mL of 6.0 M HCl (aq) 40 mL DI water + 1.0...
pH of a buffer solution pH and Buffers IV pH of a Buffer Solution A.2019 Mass of sodium acetate, CH,COONa, g Measured pH Calculated pH buffer solution prepared with dissolved CH,COONa +8.5 mL CH,COOH (aq) 5.54 40 mL buffer + 1.0 mL of 6.0 M HCI (aq) 4/.ces 40 mL buffer + 1.0 mL of 6.0 M NaOH (aq) 5.02 Calculated pH Measured pH 5,54 deionized water 40 mL DI water +1.0 mL of 6.0 M HCI (aq) 1.C03 40...
find pH is this right im confusing myself a bit we ? for calculehen Calculated pH Measured pH 4.54 deionized water should be 7 40 mL DI water + 1.0 mL of 6.0 M HCl(aq) 40 mL DI water + 1.0 mL of 6.0 M NaOH(aq) 1.57 11.93 Compare the change in pH you observed when 1.0 mL of 6.0 M HCl is added to 40 mL of the buffer versus 40 mL of water. Is there a difference, and...
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